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2302-practice-final-soln

# So c = 0 and thus y = dte − 3 t taking a derivative

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Unformatted text preview: so C = 0, and thus y = Dte − 3 t . Taking a derivative, we see that y ′ = De − 3 t- 3 Dte − 3 t . Plugging t = 0 into this equation, we get that y ′ (0) = De- 3 D e = D = 2 , so D = 2. We now have the complete solution to this initial value problem, y ( t ) = 2 te − 3 t . MAP 2302, Fall 2011 — Final Exam Review Problems 14 Now we just need to compute the maximum displacement. We know that this will occur either at t = 0 or at the first critical point. However, it cannot occur when t = 0, because y (0) = 0, so it must happen at the first critical point. Setting the first derivative equal to 0 we have y ′ = 2 e − 3 t- 6 te − 3 t = 0 . Now we can divide through by e − 3 t to see that 2- 6 t = 0, or equivalently, t = 1 / 3 . Therefore the maximum displacement is y ( 1 / 3 ) = 2 3 e − 1 . MAP 2302, Fall 2011 — Final Exam Review Problems 15 Formulas of possible use sin2 x = 2sin x cos x L { t n } = n ! s n +1 cos 2 x = cos 2 x- sin 2 x L braceleftbig e at bracerightbig = 1 s- a sin 2 x = 1- cos 2 x 2 L { sin bt } = b s 2 + b 2 cos 2 x = 1 + cos 2 x 2 L { cos bt } = s s 2 + b 2 sin( x + y ) = sin x cos y + cos x sin y L braceleftbig e at f ( t ) bracerightbig = F ( s- a ) , where F ( s ) = L { f } . cos( x + y ) = cos x cos y- sin x sin y L { tf ( t ) } =- d ds L { f } integraldisplay tan udu =- ln | cos u | L braceleftbig f ′ ( y ) bracerightbig = s L { f } - f (0) integraldisplay cot udu = ln | sin u | L { u ( t- a ) f ( t ) } = e − as L { f ( t + a ) } integraldisplay sec udu = ln | sec u + tan u | L { f , period T } = L { f T } 1- e − sT integraldisplay csc udu = ln | csc u- cot u | , L { δ ( t- a ) } = e − as sin x = ∞ summationdisplay n =0 (- 1) n x 2 n +1 (2 n + 1)! cos x = ∞ summationdisplay n =0 (- 1) n x 2 n (2 n )! e x = ∞ summationdisplay n =0 x n n ! arcsinh x = ∞ summationdisplay n =0 (- 1) n (2 n )! 4 n ( n !) 2 (2 n + 1) x 2 n +1 Variation of parameters ay ′′ + by ′ + cy = f ( t ) has solution y ( t ) = C ( t ) y 1 ( t ) + D ( t ) y 2 ( t ) where C = integraldisplay- fy 2 a ( y 1 y ′ 2- y 2 y ′ 1 ) dt D = integraldisplay fy 1 a ( y 1 y ′ 2- y 2 y ′ 1 ) dt...
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so C = 0 and thus y = Dte − 3 t Taking a derivative we...

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